Question: Find $\tan\left(75^\circ\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2+\sqrt{6}}{2-\sqrt{6}}$ (Choice B) B $\dfrac{3+\sqrt{3}}{3-\sqrt{3}}$ (Choice C) C $\dfrac{\sqrt{3}+\sqrt{2}}{8}$ (Choice D) D $\dfrac{\sqrt{3}-1}{8}$
Solution: The strategy First, we should rewrite the given angle $75^\circ$ as the sum or difference of two special angles. Then, we can use the tangent addition or subtraction identities in order to evaluate $\tan\left(75^\circ\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $75^\circ$ We can rewrite $75^\circ$ as follows. $\begin{aligned}75^\circ&=30^\circ+45^\circ\end{aligned}$ In other words, $75^\circ$ is the sum of the special angles $30^\circ$ and $45^\circ$. Evaluating $\tan\left(75^\circ\right)$ Using the tangent addition identity, we get the following. $\begin{aligned} \tan\left(75^\circ\right)&= \tan\left(30^\circ+45^\circ\right) \\\\\\ &= \dfrac{\tan\left(30^\circ\right)+\tan\left(45^\circ\right)}{1-\tan\left(30^\circ\right)\tan\left(45^\circ\right)}\\\\\\ &= \dfrac{\left(\dfrac{\sqrt{3}}{3}\right)+\left(1\right)}{1-\left(\dfrac{\sqrt{3}}{3}\right)\left(1\right)}\\\\\\ &=\dfrac{3+\sqrt{3}}{3-\sqrt{3}} \end{aligned}$ Summary $\tan\left(75^\circ\right) = \dfrac{3+\sqrt{3}}{3-\sqrt{3}}$